Question

In an experiment, to determine the percentage purity of the hydrochloric acid, 25cm3 of the acid solution made by dissolving 25.5g of impure acid in a liter of solution was found to neutralize 25cm3 of 0.4M NaOH solution. Calculate

a. Concentration of the acid in

(i) g/dm3

(ii) mol/dm3

b. percentage purity of the acid

c. percentage impurity of the acid

d. What mass of impurities contained in 25.5g?

Answer 1

V₁S₁=V₂S₂        Here, V₁= 25 cm³, V₂=25 cm³, S₂= 0.4 M

S₁ = (25×0.4) /25 M

    =0.4 M

Strength of the acid is 0.4 mol / dm³

m = n×M              Here, n= 0.4 mol / dm³, M= 36.46 g/mol

  = 0.4×36.46 g / dm³

   =14.584 g / dm³

a)

i) Concentration of the acid is 14.584 g / dm³ (Answer)

ii) Concentration of the acid is 0.4 mol / dm³ (Answer)

b)

percentage purity of the acid = (14.584/25.5) ×100

                                             = 57.19 percent (Answer)

c)

percentage impurity of the acid = 100-57.19 = 42.81 percent (Answer)

d)

mass of impurities contained in 25.5g = 25.5-14.584 g

                                                                  =10.916 g (Answer)